Difference between revisions of "2016 AMC 8 Problems/Problem 10"
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| + | ==Problem== | ||
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Suppose that <math>a * b</math> means <math>3a-b.</math> What is the value of <math>x</math> if | Suppose that <math>a * b</math> means <math>3a-b.</math> What is the value of <math>x</math> if | ||
<cmath>2 * (5 * x)=1</cmath> | <cmath>2 * (5 * x)=1</cmath> | ||
<math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math> | <math>\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14</math> | ||
| − | ==Solution== | + | ==Solutions== |
| + | |||
| + | ===Solution 1=== | ||
Let us plug in <math>(5 * x)=1</math> into <math>3a-b</math>. Thus it would be <math>3(5)-x</math>. Now we have <math>2*(15-x)=1</math>. Plugging <math>2*(15-x)</math> into <math>3a-b</math>, we have <math>6-15+x=1</math>. Solving for <math>x</math> we have <cmath>-9+x=1</cmath><cmath>x=\boxed{\textbf{(D)} \, 10}</cmath> | Let us plug in <math>(5 * x)=1</math> into <math>3a-b</math>. Thus it would be <math>3(5)-x</math>. Now we have <math>2*(15-x)=1</math>. Plugging <math>2*(15-x)</math> into <math>3a-b</math>, we have <math>6-15+x=1</math>. Solving for <math>x</math> we have <cmath>-9+x=1</cmath><cmath>x=\boxed{\textbf{(D)} \, 10}</cmath> | ||
| − | ==Solution 2== | + | ===Solution 2=== |
Let us set a variable <math>y</math> equal to <math>5 * x</math>. Solving for y in the equation <math>3(2)-y=1</math>, we see that y is equal to five. By substitution, we see that <math>5 * x</math> = 5. Solving for x in the equation <math>5(3)-x = 5</math> we get <cmath>x=\boxed{\textbf{(D)} \, 10}</cmath> | Let us set a variable <math>y</math> equal to <math>5 * x</math>. Solving for y in the equation <math>3(2)-y=1</math>, we see that y is equal to five. By substitution, we see that <math>5 * x</math> = 5. Solving for x in the equation <math>5(3)-x = 5</math> we get <cmath>x=\boxed{\textbf{(D)} \, 10}</cmath> | ||
{{AMC8 box|year=2016|num-b=9|num-a=11}} | {{AMC8 box|year=2016|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:54, 16 January 2021
Contents
Problem
Suppose that 
means 
What is the value of 
if
![\[2 * (5 * x)=1\]](http://latex.artofproblemsolving.com/c/2/f/c2f6515e388e34fda66b684dcf2dc3b67ef1bc83.png)
Solutions
Solution 1
Let us plug in 
into 
. Thus it would be 
. Now we have 
. Plugging 
into , we have 
. Solving for we have ![\[-9+x=1\]](http://latex.artofproblemsolving.com/c/6/1/c61c1f668c35a5c27c69301b25294456176e3130.png)
![\[x=\boxed{\textbf{(D)} \, 10}\]](http://latex.artofproblemsolving.com/7/9/e/79ee27e8c64bd1f7b4906b292dcb3704d68e7c15.png)
Solution 2
Let us set a variable 
equal to 
. Solving for y in the equation 
, we see that y is equal to five. By substitution, we see that = 5. Solving for x in the equation 
we get
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
